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laila [671]
3 years ago
8

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/

s2 . Ignore air resistance.
A flea jumps straight up to a maximum height of 0.460m . What is its initial velocity v0 as it leaves the ground?
Physics
1 answer:
polet [3.4K]3 years ago
8 0
V o - initial velocity
v = velocity at the maximum height,
v² = v o² - 2 g h
v = 0
0 = v o² - 2 g h
v o² = 2 g h = 2 · 9.80 · 0.460
v o² = 9.052
v o = √9.052 = 3.004197 m/s ≈ 3 m/s
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Given Information:

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Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

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